Question

Ab is the diameter and ac is a chord with centre o such that anglebac = 30 . the tangent at c intersects intersects extended ab at

d. prove that bd=bc

Answer 1

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Φ Given = A circle with AB as diameter having chord AC. ∠BAC = 30°
Tangent at C meets AB produced at D.

ΦTo Prove = BC = BD

ΦConstruction = Join OC.

ΦPROOF = ........
In Δ AOC,
= OA = OC
= ∠1 = ∠BAC [Angles opposite to equal sides are equal]
= ∠1 = 30°

By angle sum property of Δ

∠2 = 180° – (30° + 30°)
= 180° – 60°
∠2 = 120°

NOW

∠2 + ∠3 = 180° (linear pair)
= 120° + ∠3 = 180°
= ∠3= 60°

AB is diameter of the circle. [Given]

the angle is a semi circle is 90°.

= ∠ACB = 90°
= ∠1 + ∠4 = 90°
= 30° + ∠4 = 90°
= ∠4 = 60°

Consider OC is radius and CD is.
tangent to circle at C.

OC ⊥ CD
= ∠OCD = 90°
= ∠4 + ∠5 (=∠BCD) = 90°
= 60° + ∠5 = 90°
= ∠5 = 30°

In ΔOCD,

∠5 + ∠OCD + ∠6 = 180° [By angle sum property of Δ]
⇒ 60° + 90° + ∠6 = 18°
⇒ ∠6 + 15° = 180°
⇒ ∠6 = 30°

NOW

In ΔBCD
∠5 = ∠6 [= 30°]
⇒ BC = CD [sides opposite to equal angles are equal.

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